Statistics – Mean and Variance – JEE Main 23 Jan 2026 Shift 2

Solution:

First, find the mid-values ($x_i$) of the classes: 6, 10, 14, 18.
Frequencies ($f_i$): 3, $\lambda$, 4, 7.
Total Frequency $N = \sum f_i = 3 + \lambda + 4 + 7 = 14 + \lambda$.

Calculate the Mean ($\mu$):
$$ \mu = \frac{\sum f_i x_i}{N} = \frac{3(6) + \lambda(10) + 4(14) + 7(18)}{14 + \lambda} $$
$$ \mu = \frac{18 + 10\lambda + 56 + 126}{14 + \lambda} = \frac{200 + 10\lambda}{14 + \lambda} $$
$$ \mu = \frac{10(20 + \lambda)}{14 + \lambda} = 10 + \frac{60}{14 + \lambda} $$

Since $\mu$ is likely an integer (or simply for calculation simplicity), $14 + \lambda$ should be a divisor of 60.
Possible values for $14+\lambda$: 15, 20, 30, 60… $\Rightarrow \lambda \in \{1, 6, 16, 46\}$.

Given Variance ($\sigma^2$) = 19:
$$ \sigma^2 = \frac{\sum f_i x_i^2}{N} – (\mu)^2 = 19 $$
$$ \frac{3(36) + \lambda(100) + 4(196) + 7(324)}{14 + \lambda} – \mu^2 = 19 $$

Let’s test $\lambda = 6$ (a likely integer value).
If $\lambda = 6$:
$$ \mu = 10 + \frac{60}{20} = 13 $$
Calculate variance with $\lambda = 6, \mu = 13$:
$$ \sigma^2 = \frac{108 + 600 + 784 + 2268}{20} – 169 $$
$$ \sigma^2 = \frac{3760}{20} – 169 = 188 – 169 = 19 $$
This matches the given variance. Thus, $\lambda = 6$ and $\mu = 13$.

Value of $\lambda + \mu$:
$$ \lambda + \mu = 6 + 13 = 19 $$

Ans. (4)